UPSC Prelims 2023·CSAT·Logical Reasoning·Deductive Logic

Let P, Q, R, S and T be five statements such that: I. If P is true, then both Q and S are true. II. If R and S are true, then T is false. Which of the following can be concluded? 1. If T is true, then at least one of P and R must be false. 2. If Q is true, then P is true. Select the correct answer using the code given below:

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  1. A1 onlyCorrect
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Explanation

Let's analyze the given statements: I. If P is true, then both Q and S are true. (P -> (Q and S)) II. If R and S are true, then T is false. ((R and S) -> not T) Now let's evaluate each conclusion: Conclusion 1: If T is true, then at least one of P and R must be false. This can be written as: T -> (not P or not R). Let's try to prove its contrapositive: If (P and R) is true, then T must be false. Assume (P and R) is true. This means P is true AND R is true. 1. If P is true, then from Statement I, (Q and S) is true. This implies S is true. 2. We now have R is true (from our assumption) and S is true (from step 1). 3. Since (R and S) is true, then from Statement II, T must be false. So, if (P and R) is true, then T is false. This is the contrapositive of Conclusion 1. Since the contrapositive is true, Conclusion 1 itself is true. Alternatively, let's assume T is true and see what follows: 1. If T is true, then 'not T' is false. 2. From Statement II, ((R and S) -> not T). Since the consequent (not T) is false, the antecedent (R and S) must also be false for the implication to be true. 3. So, if T is true, then (R and S) is false. This means (not R or not S) is true. 4. Now we need to show that (not P or not R) is true. Case A: Assume P is true. If P is true, then from Statement I, (Q and S) is true, which means S is true. We know from step 3 that (not R or not S) is true. If S is true, then (not R or false) must be true, which implies not R is true. So, if P is true and T is true, then not R is true. This makes (not P or not R) true. Case B: Assume P is false. If P is false, then (not P) is true. This automatically makes (not P or not R) true, regardless of R. In both cases, if T is true, then (not P or not R) is true. Therefore, Conclusion 1 is correct. Conclusion 2: If Q is true, then P is true. (Q -> P) From Statement I, we have P -> (Q and S). This implies P -> Q. Conclusion 2 (Q -> P) is the converse of P -> Q. The converse of a true statement is not necessarily true. Consider a scenario where P is false, Q is true, and S is true. - Statement I (P -> (Q and S)) becomes (False -> (True and True)), which is (False -> True), which is True. So Statement I holds. - Statement II ((R and S) -> not T) can also hold (e.g., if R is true, S is true, T is false). In this scenario, Q is true, but P is false. So, "If Q is true, then P is true" is false. Therefore, Conclusion 2 cannot be concluded. Based on the analysis, only Conclusion 1 is correct. The final answer is A) 1 only.
Logical Reasoning: Let P, Q, R, S and T be five statements such that: I. If P is true, then both Q and S are true. II. If R and S are true,

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